First I need to clarify that when I speak of a bet, that is not the same as a spin. A bet can consist of a single spin, or any number of spins. A bet is playing however many spins come to the ultimate resolution of a "win" or "loss", which I will explain now.

This system has nothing to do with bet selection, nor is it really a money management system. What it is is a method of betting that holds a distinct mathematical advantage over the house, even on a 00 table.

For the sake of simplicity, I will give the following example, but it doesn't matter how you choose your dozens.

If the 2nd dozen has hit, you have two options for movement - Either the spin can move to the 1st dozen (left) or the 3rd dozen (right). A bet requires however many spins it takes to move to the left or right. If I get a series of 2-2-2-2-2-2-2-2-1, that is 9 spins but only one bet.

In this example, I anticipate a move to the right, meaning I will bet with the expectation that the spin will eventually move from the 2nd dozen to the 3rd dozen. If it moves to the right (3rd dozen), it is a "win" and the bet is over. If it moves to the left (1st dozen), it is a "loss" and the bet is over. For as long as it repeats on the 2nd dozen, the bet is not complete and we continue spinning until the bet is resolved.

I will place 1 unit on the 2nd dozen and 1 unit on the 3rd dozen. Every time it lands on either, I have +1 and if it lands on 1st dozen, I am -2 units.

Here is where the math moves in our favor: The most I can possibly lose on this bet is -2 units, whereas I have unlimited winning potential for as long as it repeats on the 2nd dozen. I will be explaining the math using a single 0 table.

On the first spin, we have a 13/37 chance of losing (1st dozen + 0), a 12/37 chance of winning (3rd dozen), and a 12/37 chance of repeating and moving to the second spin of the bet. Our expected value of this first spin is -2.7%. However, 32.4% of the time, we repeat our previous dozen and move to the second spin of the bet. As we have already gained +1 unit from the repeat on the 2nd dozen, our new payout is (0+1st dozen)=-1 [-2 units offset by our previous win], 3rd dozen=+2 (win +1 plus the +1 from previous repeat), and 2nd dozen=+2 repeats and move on to the third spin of the bet.

Third spin (which happens 10.5% of the time (12/37)^2) has payouts of (0+1st dozen)=0, 3rd dozen=+3 and 2nd dozen=+3 and move to the fourth spin, etc.

In 37 spins, we would average 13 losses of -2, 12 wins of +1, and 12 spins where we move to higher levels. Of these 12, about 4 would be losses of -1, 4 would be wins of +2, and 4 would move on again. Of these, 1 would be a loss of -0, 1 would be a win of +3, and 1 or 2 would continue on to higher winnings.

By extending the odds of each level out, this method gives a positive expectation of +3.xxx units over a single 0 table, better than +1.xxx on a 0/00 table, and better than +5.xxx on no-zero tables (if you are lucky enough to find them).

As soon as a "bet" has come to an end, with either a win or a loss, you take your money and start a new "bet", for as many spins as that takes.

Messing with the math doesn't change it, it only makes you more likely to make an error, which seems like what happened here. The Derivation of this edge comes from the fact that every bet pays of as if there are 36 numbers, but there are 37 or 38.

Quote:rdw4potusthe expected value of every spin is -2.7%. If you're showing anything else on any spin, you have a math error somewhere...

HA! you fell into the trap of not recognizing that:

"when I speak of a bet, that is not the same as a spin"

Quote:colbsterWith the power of the repeats behind us, our losses could be -2, but they could also be +3 if they came after a series of repeats. Our wins could be +1, or they could be +6 after repeats. The bet is not over until we have a firm win or loss.

But the wheel doesn't know or care about the past spins. the only thing the repeats do is delay your loss and increase the average loss size. The house edge is unaffected by your method.

Quote:colbsterThe house edge only applies when we throw chips on a certain spot and spin repeatedly on the same spots.

um, no. Seriously? no...that's pretty basic...

Quote:BuzzardI know this system works. I am currently using it to bet horses.

But it only works if you change the number of the horse you bet in each race, right?

All for a house edge of 2.7%!

Do the math. (13/37) you will lose 2. (12/37) you will win 1. (12/37) you will move to level 2, which is more than +1 per spin because you have a new (13/37)*-1, (12/37)*+2, and (12/37)*+more than 2 because you have moved to the 3rd level. You have an infinite number of repeats before the bet is over, but the maximum you can lose is -2. That is it. -2. But you can win 2 or 3 or 6 or 8. It is easy to fall back on the mantra of -2.7. If you follow this through to its logical conclusion, you see that it has a positive expectation. It can be played flat, it can be played with progression. I haven't suggested a bet selection because it doesn't matter. The continuance of repeats makes this a very elegant method of beating the house advantage, which cannot possibly cost you more than 2 units while you are allowed to win as much as random wants to present you with.

Roulette wheel manufacturers worldwide have just called all of their employees back from the Christmas holidays to start work on their new triple zero wheels....

Damn!

Quote:colbsterI understand that my lack of posts on this forum automatically categorizes me as a newbie, and I accept that, as I make the same assumptions about new posters on the other forums that I frequent. I understand that if I walk up to a roulette table and place any bet, my expectation for that spin is negative by 2.7 or 5.3 percent. I have made the same jokes about the Marti or FTL or any other first time posts that aren't based in anything more concrete than "Hey, guess what I just noticed".

Do the math. (13/37) you will lose 2. (12/37) you will win 1. (12/37) you will move to level 2, which is more than +1 per spin because you have a new (13/37)*-1, (12/37)*+2, and (12/37)*+more than 2 because you have moved to the 3rd level. You have an infinite number of repeats before the bet is over, but the maximum you can lose is -2. That is it. -2. But you can win 2 or 3 or 6 or 8. It is easy to fall back on the mantra of -2.7. If you follow this through to its logical conclusion, you see that it has a positive expectation. It can be played flat, it can be played with progression. I haven't suggested a bet selection because it doesn't matter. The continuance of repeats makes this a very elegant method of beating the house advantage, which cannot possibly cost you more than 2 units while you are allowed to win as much as random wants to present you with.

Huh??

Heres the bottom line regarding numbers that repeat. Lets say you tracked the last 38 spins. Lets say in the last 38 spins, 1 and 2 came up 4 times. 3 and 4 came up 3 times. 5-6-7-8 hit 2 times. 9-10-11-12-13-14-15-16-17-18-19-20-21-22-23-24 hit once. 25-26-27-28-29-30-31-32-33-34-35-36-0-00 never hit in the last 38 spins. Lets say I bet on the number 2. It came up 4 times in the last 38 spins. What are the odds of 4 hitting again, 1/38. Lets say I bet on 26. It hasn't come up in the last 38 spins. I fugure this number is due. The Odds, 1/38. Regardless of repeats, its allways allways 1/38 chance of any number hitting. If 23 hit 5 times in a row, next spin, 1/38 chance. If 25 never hit all day, 1/38 chance next spin.

like, they are no casinos near him. And well, you know, you can never be sure of getting paid online.

If you are near a casino, or trust online casinos, you should thank this man for an early Christmas present !

Let me try this a different way, excluding the 0 just for the sake of easing the math (I know it exists! Just follow me here.)

Level 1 (First spin) gives us -2/1/1, meaning that without the 0, we would expect a grand total of 0 gain or loss over the long run

Level 2 (After 1 repeat) gives us -1/2/2, or a net of +1 average [(-1+2+2)/3]

This is consistent with our original -2/1/1 expectation, because this +1 average is the same as the +1 we expected in Level 1

Level 3 (After 2 repeats) gives us 0/3/3, or a net of +2 average [(0+3+3)/3]

While this only occurs in 1/27 of our spins, it has already changed the expected value on a no-zero table to -2/1.037/1. With a z-score well within expectations, we are already in a positive scenario on a no-zero table

Level 4 (After 3 repeats) gives us 1/4/4, or an net of +3 average, further moving our expected payout beyond that of the house edge.

I haven't even branched out to further degrees of deviation. While each level is less likely to occur, when it does, it is increasingly in our favor, further skewing the results in our favor.

There is a boat about an hour east of me with 0/00 tables, where I have enjoyed modest success, but the odds are sufficiently harsh that I'm lucky to break even in such a short duration of play and higher table minimums that I am comfortable with.

I fall asleep at night thinking numbers. I originally started running roulette numbers through my brain to help me doze off. It wasn't about the money, although I have played for money and find myself uncomfortable with the risk, even though the math is solid. I will continue to play low stakes when they are available to me.

It makes no difference that people criticize without making the effort to understand this. If you wish to PM me, I would be glad to point you to my posts on other forums where this has been discussed ad nauseum and tested by people with plenty of posts by their names, which I think is what gives superficial credibility to posters.

I bring this to the discussion because I think those who take the time to get what I am talking about will benefit, and I enjoy the discussion, even with those that disagree with me. The math is solid, I gain nothing from sharing it freely, and I don't live or die based on the support of any in this community. I love the game of roulette for its mathematical symmetry and find it beautiful. I enjoy having open discussions about the merits of ideas, and will continue to argue in defense of this method, which is the result of many others contributions to its development.

Quote:colbster

Level 1 (First spin) gives us -2/1/1, meaning that without the 0, we would expect a grand total of 0 gain or loss over the long run

Level 2 (After 1 repeat) gives us -1/2/2, or a net of +1 average [(-1+2+2)/3]

This is consistent with our original -2/1/1 expectation, because this +1 average is the same as the +1 we expected in Level 1

OK, so if I'm reading this right...in level 1, you're betting on 2 of 3 dozens with a total of 2 units wagered. if one of your bets wins, you will have 3 total units (the one you bet, plus the 2 you won). Your total bets on level 2 equal 4 units. Where'd the extra unit wagered come from? Isn't that one unit that comes out of your pocket equal to the "+1" average for this level?

You bet 1 unit on 2 dozens, in this example dozens 2 and 3. Your payout on the 3 dozens are -2/1/1 repectively.

When you get a repeat on dozen 2, you place that money to the side. You begin your spin +1 and bet 1 unit again each on the 2nd and 3rd dozens. The payout is still -2/1/1, but you started the 2nd spin +1 so the net payout would be -1/2/2. When you get a repeat, you are now +2 to the side, bet 1 unit as before on 2nd and 3rd dozens. The payout is still -2/1/1, but you start at +2 this spin, meaning your net payout would be 0/3/3, etc.

You don't ever increase the size of the bet, but your starting position moves up 1 unit with each successive repeat.

Hope that is clearer.

Quote:colbsterSorry, I guess I didn't explain myself properly.

You bet 1 unit on 2 dozens, in this example dozens 2 and 3. Your payout on the 3 dozens are -2/1/1 repectively.

When you get a repeat on dozen 2, you place that money to the side. You begin your spin +1 and bet 1 unit again each on the 2nd and 3rd dozens. The payout is still -2/1/1, but you started the 2nd spin +1 so the net payout would be -1/2/2. When you get a repeat, you are now +2 to the side, bet 1 unit as before on 2nd and 3rd dozens. The payout is still -2/1/1, but you start at +2 this spin, meaning your net payout would be 0/3/3, etc.

You don't ever increase the size of the bet, but your starting position moves up 1 unit with each successive repeat.

Hope that is clearer.

I guess that's clearer. But that just means that your expected value of each spin is 0 (-2/1/1). If you happen to get ahead temporarily, you're basically just waiting for the frequency of the first dozen hits to even out. There's no long-term expected gain, since each individual spin is expected to make you 0 (or -2.7%, if we put the 0 back in).

Quote:colbsterSorry, I guess I didn't explain myself properly.

You bet 1 unit on 2 dozens, in this example dozens 2 and 3. Your payout on the 3 dozens are -2/1/1 repectively.

When you get a repeat on dozen 2, you place that money to the side. You begin your spin +1 and bet 1 unit again each on the 2nd and 3rd dozens. The payout is still -2/1/1, but you started the 2nd spin +1 so the net payout would be -1/2/2. When you get a repeat, you are now +2 to the side, bet 1 unit as before on 2nd and 3rd dozens. The payout is still -2/1/1, but you start at +2 this spin, meaning your net payout would be 0/3/3, etc.

You don't ever increase the size of the bet, but your starting position moves up 1 unit with each successive repeat.

Hope that is clearer.

Since you have just published your system for the world to admire, I would try to selfishly exploit it for your own benefit before casinos ban it from play. I feel confident that if you write a proposal to a casino manager or a casino host and take out a line of credit they will comp you room, food and beverage. They may even send a limo to the airport to pick you up. How can you lose? Hurry before your system is banned.

Quote:colbsterEach of them moves the bar farther towards our side of the equation.

Only temporarily. So I get to level 3. OK. And a number in the first dozen rolls. and then a number in the first dozen rolls again. and again. It's just as likely that the first dozen repeats as any other dozen. It's very easy to see what the long-term expectation is, so it's also very easy to see that getting to the 4th or 5th level and "booking" those gains is temporary.

Quote:rdw4potusSo I get to level 3. OK. And a number in the first dozen rolls. and then a number in the first dozen rolls again. and again. It's just as likely that the first dozen repeats as any other dozen.

No. After the 1st dozen hits the first time, this bet is a loss. Since you are at Level 3, your net loss is 0 because you have already pocketed 2 units from the repeats, lost 2 from the bet on the table, and they offset. No harm, no foul. The bet is over now that it has been resolved, though.

Say I was anticipating that the spins would move to the right. Once your first spin of the 1st dozen came in, I moved my bet to the 1st and 2nd dozens (repeat plus the move to the right). You are correct that the first dozen can repeat, and we gain from that fact. Each bet has a positive expectation. As soon as the bet is over, you start a new bet on a new 2 dozens, chosen however you choose to select them. That new bet has a positive expectation, just like the previous.

How do you gain any information from the fact that dozens may or may not repeat at an exactly fair and random frequency?

Quote:odiousgambitall these guys couldnt prove you wrong and nobody has tested it with a simulator?

The problem I have explaining this is that it is difficult to separate the money aspect from the bet selection. If you run a simulator, you have to enter some sort of bet selection and people like to argue that if it wins, it is the bet selection, but if it loses, it is the fault of the staking method. As I said initially, this is not money management or bet selection. It is its own category which I cannot put proper words to.

I liken it to knowing the speed of light. So you know something? Great! How the heck are you going to apply it to something worthwhile. I can look up the speed of light but it will not allow me to time travel or escape gravity or anything else that sounds fun. I have to apply the knowledge to a physical action for it to benefit me. I know that this staking holds an advantage over the house. Where it can get tripped up is applying it to an actual bet selection that is neutral. Unfortunately, that is difficult. It is gambler's fallacy or relying too much on streaks. With neutral bet selection, it is a winner.

Quote:colbsterNo. After the 1st dozen hits the first time, this bet is a loss. Since you are at Level 3, your net loss is 0 because you have already pocketed 2 units from the repeats, lost 2 from the bet on the table, and they offset. No harm, no foul. The bet is over now that it has been resolved, though.

Say I was anticipating that the spins would move to the right. Once your first spin of the 1st dozen came in, I moved my bet to the 1st and 2nd dozens (repeat plus the move to the right). You are correct that the first dozen can repeat, and we gain from that fact. Each bet has a positive expectation. As soon as the bet is over, you start a new bet on a new 2 dozens, chosen however you choose to select them. That new bet has a positive expectation, just like the previous.

the odds don't change based on the dozens that you choose. you're still break-even on every bet with a no-0 wheel. There's nothing you can do to either make or lose money in the long run. Then, adding the 0s makes every bet have a negative expectation.

Quote:Boney526How do you "anticipate the spins would move to the right"?

You cannot anticipate the spins. That is exactly where my trouble comes in. I have to couple the math with a means of applying it. I can arbitrarily pick "to the right" and it wins with a perfect distribution of spins. Earlier today, I did just that and experienced a horrific streak of "to the left". If you look at what I have shared in a prism of just staking, it is amazing. Where to place the bets brings in its own challenges that cannot be ignored, but my method at least offsets the effective house advantage so we can focus on choosing the best progression and/or bet selection.

Quote:rdw4potusthe odds don't change based on the dozens that you choose. you're still break-even on every bet with a no-0 wheel. There's nothing you can do to either make or lose money in the long run. Then, adding the 0s makes every bet have a negative expectation.

No, the odds don't change. They are still 1/12/12/12 out of 37 spins. However, on your initial spin, you are at +/- 0, as you are just walking up to the table. After you hit a repeat, you are still 1/12/12/12, but you begin with a chip in your pocket that wasn't there when you walked up. As I have said, your new bet profile is no longer -2/1/1, but rather -1/2/2.

Quote:Buzzard" Then, adding the 0s makes every bet have a negative expectation. "

No, adding the 0s in makes every "spin" have a negative expectation. I have explained that this deals with "bets", not "spins". When "bets" can contain a number of "spins", you no longer have the negative expectation.

Quote:colbsterNo, the odds don't change. They are still 1/12/12/12 out of 37 spins. However, on your initial spin, you are at +/- 0, as you are just walking up to the table. After you hit a repeat, you are still 1/12/12/12, but you begin with a chip in your pocket that wasn't there when you walked up. As I have said, your new bet profile is no longer -2/1/1, but rather -1/2/2.

Unless you lose first. Then you begin with 2 fewer chips in your pocket. or 4 fewer chips. or 6 fewer chips, or.......

Until then you will just be another fool, in case you had not noticed !

Quote:rdw4potusUnless you lose first. Then you begin with 2 fewer chips in your pocket. or 4 fewer chips. or 6 fewer chips, or.......

You can only end up with 2 fewer chips, then it all starts again at 0. This is where my system is different than you might expect. It goes back to the definition of "bet" and "spin". Like any other system, you will lose plenty of times and you will win plenty of times. The difference comes from letting the repeats run, pocketing money along the way as they do.

Quote:BuzzardSince you are new to this forum, you might not know EvenBob is our resident Roulette expert. You can click on his name or search threads on Roulette. Or PM him. No one is going to take you serious until EvenBob has commented on your system.

Until then you will just be another fool, in case you had not noticed !

I get that. My every day forum has a couple guys that give their "blessing" to a system and it gets legs. If I was worried about being perceived as a fool, I wouldn't have spent a good part of my day responding to posts and some PMs. EvenBob can opine if he chooses. I hope he does, because I always appreciate well-thought-out feedback from those who have earned the respect of the community. I don't expect to have that respect today, or tomorrow. That doesn't change the merits of the system I have outlined.

If Bob sees the thread, he can say what he chooses. I am certain that I will enjoy conversing with him about this method as I have with numerous others elsewhere.

Quote:colbsterYou can only end up with 2 fewer chips, then it all starts again at 0. This is where my system is different than you might expect.

So you cumulate your winnings, but discard your losses. Great system.

Quote:MangoJSo you cumulate your winnings, but discard your losses. Great system.

Only within the realm of a single bet. Numerous bets will end with anywhere between -2 and +??? each, aggregating into a larger negative or positive number. I have had some fair size losses and some fair size wins, as the reality of the table is never 1/12/12/12. The point I am making is that this is not -2/1/1 as some expect. It works out to -2/1.???/1 and gives us an advantage at the end of the day. If you have a silly bet selection, you will lose. With a great selection, you could also still lose when random decides she is angry at you. But in the pure confines of probability, this it better than break even.

Quote:colbsterYou can only end up with 2 fewer chips, then it all starts again at 0. This is where my system is different than you might expect. It goes back to the definition of "bet" and "spin". Like any other system, you will lose plenty of times and you will win plenty of times. The difference comes from letting the repeats run, pocketing money along the way as they do.

How can you only end up with 2 fewer chips? How does it start again at 0? Didn't you just lose 2 chips? How are you not at -2 cumulatively at that point? If I lose twice in a row, I'm down 4 on the day. I can't magically choose to be down "only" 2 after the second loss.

Quote:colbsterBut in the pure confines of probability, this it better than break even.

nope. clearly not.

Quote:rdw4potusHow can you only end up with 2 fewer chips? How does it start again at 0? Didn't you just lose 2 chips? How are you not at -2 cumulatively at that point? If I lose twice in a row, I'm down 4 on the day. I can't magically choose to be down "only" 2 after the second loss.

The same way that you don't want to acknowledge that I might start the next round of bets at +8. As we all can agree, the table does not have a memory. Each bet is a new bet that begins at 0. Now, you may string together a series of bets that don't begin or end at 0, but on this bet, right this second, the table views you as break even. You can only lose a bet once, losing a maximum of 2 units. The next spin is the beginning of the next bet, which start at 0, can drop to -2, or can grow to any positive number you are willing to accept as statistically possible.

These bets will grow shrink your bankroll, and there is always the chance you will leave the table with less than when you showed up. Each "bet" has a mathematical advantage because you limit your losses but keep your wins limitless.

Quote:colbsterEach "bet" has a mathematical advantage because you limit your losses but keep your wins limitless.

Dude, seriously, NO. No amount of bankroll management does anything at all (even a little bit) to change the fact that the game has a negative expectation on every spin.